mathematics in the modern world, Assignments of Mathematics

Download mathematics in the modern world and more Assignments Mathematics in PDF only on Docsity! Continuity The function f(x) when x approches a is continuous if this 3 condition are satisfied 1. f(a) is exist 2. lim 𝑥 𝑎 (𝑓𝑥) 𝑒𝑥𝑖𝑠𝑡 3. lim 𝑥 𝑎 (𝑓𝑥) = 𝑓(𝑎) Example: At x=2, f(x)=𝑥2 + 1 For Condition 1 f(a) is exist F(2)= (22) + 1 = 5 For condition 2 lim 𝑥 𝑎 𝑓(𝑥) We have lim 𝑥 2 (𝑥2 + 1) = lim 𝑥 2 𝑥 . lim 𝑥 2 𝑥 + lim 𝑥 2 1 = 2.2 + 1 = 5 Then the last condition w/c is f(a)=lim 𝑥 2 (𝑥2 + 1) 5=5 so the problem i continuous because it satisfy all the 3 condition If any of the condition is not satisfied then the problem is discontinuous 2 kind of Discontinuity 1st Kind if Left and Right hand Limit exist If lim 𝑥 𝑎+ 𝑓(𝑥) = lim 𝑥 𝑎− 𝑓(𝑥) is called removable discontinuity If lim 𝑥 𝑎+ 𝑓(𝑥) ≠ lim 𝑥 𝑎− 𝑓(𝑥) is called jump discontinuity (Finite Jumps) 2nd Kind is called non removable or essential discontinuity(Infinite Discontinuities) If one of the one side limit does not exist of infinite Example: Investigate continuity of the function y= 3 𝑥 1−𝑥2 if we let x=±1 the function will be undefine so the function has discontinuity at x=±1 we have to find the left and right hand limit at the point of discontinuity to define what kind of discontinuity is the function at right hand limit @ 1 lim 𝑥 1+ 3 𝑥 1−𝑥2 = 3 lim 𝑥 1+ 𝑥 lim 𝑥 1+ 1− lim 𝑥 1+ 𝑥. lim 𝑥 1+ 𝑥 = 3 1 1−1.1 = 3 1 0 = 3∞ = 0 Left hand limit @ x=1 lim 𝑥 1− 3 𝑥 1−𝑥2 = 3 lim 𝑥 1− 𝑥 lim 𝑥 1− 1− lim 𝑥 1− 𝑥. lim 𝑥 1− 𝑥 = 3 1 1−1.1 = 3 1 −0 = 3−∞ = 1 3∞ = ∞ It is essential discontinuity because one of the one side limit is infinite Example: as x approches to 1, y= 1 (𝑥−1)2 Solution: lim 𝑥 1 1 (𝑥 − 1)2 = lim 𝑥 1 1 ((lim 𝑥 1 𝑥 − lim 𝑥 1 1)2 = 1 (1 − 1)2 = 1 0 = ∞ At x = 1 the function has Infinite Discontinuities Function with a Approches Infinity Example : lim 𝑥 ∞ 𝑥2−1 𝑥2+1 Solution: lim 𝑥 ∞ 𝑥2 − 1 𝑥2 + 1 = lim 𝑥 ∞ 𝑥2 𝑥2 − 1 𝑥2 𝑥2 𝑥2 + 1 𝑥2 = lim 𝑥 ∞ 1 − lim 𝑥 ∞ 1 𝑥2 lim 𝑥 ∞ 1 + lim 𝑥 ∞ 1 𝑥2 = 1 − 0 1 + 0 = 1 Example: lim 𝑥 ∞ 𝑥2 𝑥−2 Solution: lim 𝑥 ∞ 𝑥2 𝑥−2 = lim 𝑥 ∞ 𝑥2 𝑥 𝑥 𝑥 − 2 𝑥 = lim 𝑥 ∞ 𝑥 lim 𝑥 ∞ 1−lim 𝑥 ∞ 2 𝑥 =0